Download MATH 2105 HOMEWORK SET 3, SOLUTIONS Problem 11 (3.4(34

MATH 2105 HOMEWORK SET 3, SOLUTIONS
Problem 11 (3.4(34)): Reorgnize the following statements to obtain a coherent conclusion.
(1)
(2)
(3)
(4)
(5)
All writers who understand human nature are clever.
No one is a true poet unless he can stir the human heart.
Shakespeare wrote Hamlet.
No one who doesn’t understand human nature can stir the human heart.
None but a true poet could have written Hamlet.
It us helpful to set the domain of discourse to consist of the writers, and then to identify
the logical and the nonlogical terms in these statements: From sentence (1), let U (x), C(x)
say “x understands human nature (resp., is clever).” From sentence (2), let P (x), S(x) say
“x is a true poet (resp., can stir the human heart.” From sentence (3), let s be a name
for Shakespeare and let H(x) say “x wrote Hamlet.” Sentences (4,5) don’t add any new
predicates, so let’s paraphrase the statements using the notion of predicate logic. Sentence
(1) easily translates to ∀x[U (x) → C(x)]. Sentence (2) needs to be analyzed a little bit;
it basically says that if you can’t stir the human heart, you’re not a true poet. This then
becomes ∀x[∼ S →∼ P ]; equivalently, ∀x[P (x) → S(x)]. Sentence (3) is simply H(s). Sentence (4) becomes ∀x[∼ U (x) →∼ S(x)]; equivalently, ∀x[S(x) → U (x)]. Finally, sentence
(5) says anyone who wrote Hamlet had to have been a true poet; i.e., ∀x[H(x) → P (x)]. So
now we list these translated statements:
(1’)
(2’)
(3’)
(4’)
(5’)
∀x[U (x) → C(x)]
∀x[P (x) → S(x)]
H(s)
∀x[S(x) → U (x)]
∀x[H(x) → P (x)]
From (3’) and (5’), we conclude P (s); from that and (2’) we conclude S(s). Next, using
(4’), we have U (s); finally from (1’) we have C(s). So Shakespeare was clever, assuming all
the hypotheses true.
Problem 12 (4.1(52)): Prove or refute: “For every integer m > 2, m2 − 4 is composite.”
Noting that m2 − 4 = (m − 2)(m + 2), one is tempted to say this statement is true. However,
both factors have to be different from m2 − 4 as well as 1, and this gives us pause: what if
m = 3? Then m − 2 = 1. Not only that, m2 − 4 = 5, which is prime. So the statement is
false because we have a counterexample. But it’s the only counterexample; if we’d amended
the statement to require m > 3, it would then be true.
Problem 13 (4.2(30)): If b, c ∈ Q and one solution of x2 + bx + c = 0 is rational, then so is the other solution.
If r, s are the solutions to x2 + bx + c = 0, then–by highschool algebra–we have x2 + bx + c =
(x − r)(x − s) holding for all values of x. When we multiply the right side out, we have
x2 − (r + s)x + rs; in particular, we have r + s = −b. (If two polynomials are identically
equal, then they’re of the same degree and corresponding coefficients are equal.) If we assume, say, r ∈ Q, then s = −b − r ∈ Q as well, because b ∈ Q and Q is closed under sums
and negation.
1
2
Problem 14 (4.4(24)): If m mod 5 = 2 and n mod 3 = 6, show mn mod 5 = 1.
This problem makes no sense because n mod 3 has to be either 0,1,or 2. So this is a BOGUS
problem and will not be graded. [For a variation, you could show that mn mod 3 = 0 from
m mod 5 = 2 and n mod 6 = 3.]
Problem 15 (4.6(28)): Prove using argument by contraposition and argument by contradiction:
For all integers m, n, if mn is even, then either m is even or n is even.
For fixed integers m, n, let P be the statement “mn is even,” and let Q be the statement
“either m is even or n is even.” To prove P → Q by contraposition, we assume ∼ Q and
try to show ∼ P . ∼ Q is the statement “m is not even and n is not even,” and the parity
property tells us that this is equivalent to saying that both m and n are odd. Thus there
exist integers k, l such that m = 2k + 1 and n = 2l + 1; so mn = (2k + 1)(2l + 1) =
4kl + 2k + 2l + 1 = 2(2kl + k + l) + 1, an odd number. This is the negation of P , as desired.
As for proof by contradiction, we assume P → Q is false and try to prove a contradiction.
∼ (P → Q) is equivalent to P ∧ ∼ Q; i.e., that mn is even and that both m and n are odd.
Then the conclusion that mn is odd—as in the last paragraph–contradicts P .